(i) Assume that \(\mu \)  is a 1-logarithmic 2-Carleson measure and let \(d\nu (t)=\log \frac{2}{1-t}d\mu (t)\) . Using Proposition 2.5 of [1]}, we obtain that \(\nu \)  is a 2-Carleson measure.
Applying (REF ) and Proposition REF , we have
\(\begin{split}\int _{\mathbb {D}} \overline{I_{\mu ,\alpha }(f)(r z)} g(r z) (1-|z|^2)^{\alpha -2}d A(z)=&\int _{[0,1)} \overline{f(t)} g(r^2t) d \mu (t)\\\le & C\Vert f\Vert _{{B}} \int _{[0,1)}|g(r^2t)| \log \frac{2}{1-t} d \mu (t) \\=&C\Vert f\Vert _{{B}} \int _{[0,1)}|g(r^2t)| d\nu (t)\\\le & C\Vert f\Vert _{{B}}\Vert g\Vert _{A^1},\quad f\in {B},~g\in A^1.\end{split}\) 

Then (REF ) shows that \(\mathcal {I}_{\mu ,\alpha }\)  is a bounded operator from \({B}\)  into \({B}_{\alpha -1}\) .
Conversely, suppose that \(\mathcal {I}_{\mu ,\alpha }\)  is a bounded operator from \({B}\)  into \({B}_{\alpha -1}\) . By (REF ), we have
\(\left|\int _{[0,1)} \overline{f(t)} g(r^2t) d \mu (t)\right|\le C\Vert f\Vert _{{B}}\Vert g\Vert _{A^1},\quad 0<r<1,~f\in {B},~g\in A^1.\) 

For \(0<a<1\) , we set
\(f_{a}(z)=\log \frac{e}{1-a z}~\text{and}~ g_{a}(z)=\left(\frac{1-a^{2}}{(1-a z)^{2}}\right)^{2}, \quad z \in \mathbb {D}.\) 
Then
\(\sup _{0<a<1}\Vert f_a\Vert _{{B}}\le 3,\quad \sup _{0<a<1}\Vert g_a\Vert _{A^1}=1.\) 

Taking \(r\in [a,1)\) , we deduce that
\(\infty &>C\sup _{0<a<1}\Vert f_a\Vert _{{B}}\sup _{0<a<1}\Vert g_a\Vert _{A^1}\\& \geqslant \left|\int _{[0,1)} f_{a}(t) g_{a}(r^2t) d \mu (t)\right| \\& \geqslant \int _{a}^{1}\left(\frac{1-a^{2}}{(1-ar^2 t)^{2}}\right)^{2} \log \frac{e}{1-a t} d \mu (t) \\& \geqslant C_1\frac{\log \frac{e}{1-a^{2}}}{\left(1-a^{2}\right)^{2}} \mu ([a, 1)).\) 

This implies that \(\mu \)  is a 1-logarithmic 2-Carleson measure.
(ii) First suppose that \(\mu \)  is a vanishing 1-logarithmic 2-Carleson measure. Let \(\lbrace f_n\rbrace _{n=1}^{\infty }\)  be a bounded sequence in the Bloch space which converges to 0 uniformly on any compact subset of \(\mathbb {D}\) . Then
\(\lim _{n\rightarrow \infty }\int _{[0, r)}\left|f_{n}(t)\right||g(t)| d \mu (t)=0,\quad g\in A^1.\) 

Also, let \(d\nu (t)=\log \frac{2}{1-t}d\mu (t)\) . Then \(\nu \)  is a vanishing 2-Carleson. For \(0<r<1\) , we set
\(d\nu _r(z)=\chi _{r<|z|<1}(t)d\nu (t).\) 
Therefore,
\(\int _{[r, 1)}\left|f_{n}(t)\right||g(t)| d \mu (t)\le \int _{[0,1)}|g(t)| d \nu _{r}(t)<C \mathcal {N}\left(\nu _{r}\right)\Vert g\Vert _{A^{1}},\quad g\in A^1,\) 

where \(\mathcal {N}(\mu )\)  is the norm of identity mapping \(i\)  from \(A^1\)  into \(L^1(\mathbb {D},~\mu )\) . It is well known that \(\mathcal {N}\left(\nu _{r}\right)\rightarrow 0\)  as \(r\rightarrow 1^{-}\)  if and only if \(\nu \)  is a vanishing 2-Carleson measure. This together with (REF ), (REF ) imply that
\(\lim _{n \rightarrow \infty } \int _{[0,1)}\left|f_{n}(t) \Vert  g(t)\right| d \mu (t)=0, \quad \text{ for all } g \in A^{1}.\) 
Hence Proposition REF  shows
\(\lim _{n \rightarrow \infty }\left(\lim _{r \rightarrow 1} \left|\int _{\mathbb {D}} \overline{I_{\mu ,\alpha }(f_{n}) (rz)} g\left(rz\right) (1-|z|^2)^{\alpha -2}d A(z) \right|\right)=0, \quad \text{ for all } g \in A^{1}.\) 
Thus by (REF ), it follows that \(\mathcal {I}_{\mu ,\alpha }(f_n)\rightarrow 0\)  in \({B}_{\alpha -1}\) . So \(\mathcal {I}_{\mu ,\alpha }\)  is a compact operator from \({B}\)  into \({B}_{\alpha -1}\) .
Suppose now that \(\mathcal {I}_{\mu ,\alpha }\)  is a compact operator from \({B}\)  into \({B}_{\alpha -1}\) . Let \(\lbrace a_n\rbrace \subset (0,1)\)  be any sequence with \(a_n\rightarrow 1\) . We set
\(f_{a_n}(z)=\frac{1}{\log \frac{2}{1-a_n^2}}\left(\log \frac{2}{1-a_{n}z}\right)^2.\) 
A calculation shows that \(f_{a_n}\in {B}\) , \(\sup _{n\ge 1}\Vert f_{a_n}\Vert _{{B}}<\infty \)  and \(\lbrace f_{a_n}\rbrace \)  is bounded uniformly sequence in
\({B}\)  which converges to 0 uniformly on every compact subset of \(\mathbb {D}\) . Then \(\lbrace \mathcal {I}_{\mu _2}(f_{a_n})\rbrace \)  converges to 0 in \({B}_{\alpha -1}\) . We obtain that for every \(g\in A^1\) ,
\(\begin{split}&\lim _{n\rightarrow \infty }\int _{[0,1)} \overline{f_{a_n}(t)} g_{a_n}(r^2t) d \mu (t)\\=&\lim _{n\rightarrow \infty }\int _{\mathbb {D}} \overline{I_{\mu ,\alpha }(f_{a_n})(rz)}g_{a_n}(rz)(1-|z|^2)^{\alpha -2}d A(z)=0,\quad 0<r<1.\end{split}\) 

We also set
\(g_{a_n}(z)=\left(\frac{1-a_n^{2}}{(1-{a_n}z)^{2}}\right)^{2}\in A^1.\) 
For \(r\in (a_n,1),\)  we deduce that
\(&\int _{[0,1)} \overline{f_{a_n}(t)} g_{a_n}(r^2t) d \mu (t) \\\ge & \int _{a_n}^{1}\left(\frac{1-a_n^{2}}{(1-{a_n} r^2t)^{2}}\right)^{2} \frac{1}{\log \frac{2}{1-a_n^2}}\left(\log \frac{2}{1-a_{n}t}\right)^2 d\mu (t) \\\ge & C\frac{\log \frac{2}{1-a_n^{2}}}{\left(1-a_n^{2}\right)^{2}} \mu ([a_n, 1)).\) 

Letting \(n\rightarrow \infty \)  and bearing in mind that \(\lbrace a_n\rbrace \subset (0,1)\)  is a sequence with \(a_n\rightarrow 1\) , (REF ) gives
\(\lim _{a\rightarrow 1^-}\frac{\log \frac{2}{1-a}}{\left(1-a\right)^{2}} \mu ([a, 1))=0,\) 
which is equivalent to saying that \(\mu \)  is a vanishing 1-logarithmic 2-Carleson measure. This completes the proof.
Theorem 2.3 
For \(\alpha \ge 2\) , \(\beta >1\) , if \(\mu \)  is a positive Borel measure with \(\int _{[0,1)}\frac{1}{(1-t)^{\beta -1}}d\mu (t)<\infty .\)
\((i)\)
\(\mathcal {I}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \({B}_{\alpha -1}\)  if and only if \(\mu \)  is a (\(\beta \) +1)-Carleson measure.
\((ii)\)
\(\mathcal {I}_{\mu ,\alpha }\)  is a compact operator from \({B}\)  into \({B}_{\alpha -1}\)  if and only if \(\mu \)  is a vanishing (\(\beta \) +1)-Carleson measure.


